Author Topic: Iodination of Vanillin  (Read 15422 times)

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moo

  • Guest
I think only phenols can be iodinated that...
« Reply #20 on: March 15, 2004, 11:28:00 PM »
I think only phenols can be iodinated that way.


Mountain_Girl

  • Guest
KI or I2 ?
« Reply #21 on: May 04, 2004, 12:11:00 PM »
Rhodium's post :

5-Iodovanillin is prepared by dissolving vanillin in dilute sodium hydroxide and to that a solution of KI3 (iodine tincture?) is added dropwise. When the reaction is over, excess iodine is reduced with bisulfite, the precipitate filtered and recrystallized from GAA to give colorless prisms of 5-iodovanillin in 95% yield, mp 181.5°C.




I'm a bit confused cos the original text says:

"3.04 g Vanillin werden in 10 cc 0.2 n Natronlauge gelost, hierauf wird mit Wasser auf 25 cc verdunt.
5.08 g Jod werden dann in 50 cc Wasser durch Zusatz der notigen Menge Kaliumjodid gelost und diese Losung unter Ruhren langsam zu der Vanillinnatriumlosung gegeben. Das Jod entfarbt sich schnell und gegen Ende der Reaktion fallt ein Brei von 5-Jodvanillin aus."

My german is extremely basic but does this mean 'dissolve 5.08 g Pottasium Iodide in 50 cc water and add to the Sodium Vanillate solution' ?
Why do they refer to '5.08 g Jod' ?




hypo

  • Guest
no
« Reply #22 on: May 04, 2004, 05:17:00 PM »
"5.08g iodine are dissolved in 50ml water by adding the needed amount of KI
and this solution is slowly added to the Na-vanillate solution".

hth


Mountain_Girl

  • Guest
Iodovanillin from tincture..but not too much
« Reply #23 on: June 26, 2004, 03:45:00 PM »
Ah thanks Hypo.

Wuz wondering..
tincture is expensive & comes in a fixed concentration of 2.5 % KI, 2.5% I2 (in my case),
KI is required to dissolve the I2,
would it be feasible to add to the Na vanillate soln your tincture,
then as the dissolved I2 reacts (the 2.5% originally present),
solid I2 is added to the mixture (recovered from tincture naturally) at the same rate that it is used up so that there is always enough KI to dissolve all of it (i.e. maintain less than or equal to about 2.5% I2) until all the required I2 is added...

Hmmm..hope I has made a sensible question...wine is fucking with me receptors and my reception so its hard to tell